Solve The Laws of Motion: Tension in a Bird on a Wire

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SUMMARY

The discussion focuses on calculating the tension in a wire when a 1.00-kg bird lands on it, causing a sag of 0.200 m between two telephone poles spaced 50.0 m apart. The weight of the bird is determined to be 9.8 N, which must be balanced by the vertical components of the tension in the wire. Each tension component is calculated to be 4.9 N, as the tensions on both sides of the wire are equal. The problem emphasizes the importance of drawing a free-body diagram to visualize the forces acting on the bird and the wire.

PREREQUISITES
  • Understanding of Newton's laws of motion, specifically F=ma
  • Ability to draw and interpret free-body diagrams
  • Knowledge of vector components, particularly in tension scenarios
  • Basic principles of static equilibrium in physics
NEXT STEPS
  • Explore the concept of static equilibrium in more detail
  • Learn how to resolve forces into horizontal and vertical components
  • Study examples of tension in cables and wires in physics problems
  • Investigate the effects of different weights and angles on tension calculations
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Students studying physics, particularly those focusing on mechanics and the laws of motion, as well as educators looking for practical examples of tension in real-world applications.

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1. The distance between two telephone poles is 50.0 m. When
a 1.00-kg bird lands on the telephone wire midway between
the poles, the wire sags 0.200 m. (a) Draw a free-body diagram
of the bird. (b) How much tension does the bird produce
in the wire? Ignore the weight of the wire.2. this problem was from the laws of motion section. dealing with tension and f=ma
3. i need help with beginning the problem. any suggestions.so i drew the free body diagram. so would t1 = t2 make the tension the same for both sides? and i found the weight of the bird which is 9.8 N. so where do i go from here to find the tension?
 
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You have enough information to make a sketch of the situation--do that first. Since it is a massless wire, assume initially that the unloaded wire is perfectly straight with no initial tension.
 
After you draw the FBD, the weight of the bird would cause the wire to drop a bit and create an angle from the original horizontal position.

Yes, the tensions are exactly the same. But you first need to find the horizontal and vertical components of the tensions (you only have to do this once because the tensions are the same).

In the vertical direction, you know that the bird is exerting a force downwards equal to 9.8N, and so the tensions of the wires must add together to exert a force upwards equal to 9.8N. Because there are two separate tensions, each tension has a vertical component of 4.9N.

See how far you can get with this.
 

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